Infer mean of multivariate r.v. by observing biased subset of points Say you have a multivariate normal r.v. $X$ with unknown mean $\mu_X \in \mathbb{R}^n$ and known covariance matrix $\Sigma_x \in\mathbb{R}^{n\times n}$. You want to estimate the mean $\mu_X$ but you only observe a subset of the r.v. $Y\subset X$ defined by $Y=\{x\in X\mid x^T\Sigma^{-1}x>\lambda\}$.
In other words, if $Z$ is the multivariate normal r.v. with covariance matrix $\Sigma_x$ and mean $0_n$, then $Y$ is the set of points in $X$ that have a Mahalanobis distance to $Z$ which is greater than $\sqrt{\lambda}$, where $\lambda\in\mathbb{R}^+$ and $\lambda$ is known.
If you could observe all the points in $X$, you could simply take their average to get an estimate for the mean $\mu_X$. But your sample is biased by this minimum Mahalanobis distance to $Z$ requirement. It seems to me like you should be able to estimate $\mu_X$ since you know exactly how your sample is biased, but I am struggling to see how.
Using this information, can you estimate $\mu_X$? If so, how? And if not, what other information would you need in order to estimate it?
Visual Example:
Here's a visual of what I'm talking about. In this example, all the green points would be the from the distribution $X$, but only the dark green points are observed (they are the points from the distribution $Y$). The black ellipse is the boundary representing the minimum Mahalanobis distance to $Z$. And I want to estimate the mean of $X$, shown here with a blue plus. It seems like there should be enough information to do this but I am not quite sure how.

 A: We have that
$$\mu=E[X_j]=E[X_j\mathbb{I}_{x^T\Sigma^{-1}x>\lambda}(X_j)]+\underbrace{E[X_j\mathbb{I}_{x^T\Sigma^{-1}x\leq\lambda}(X_j)]}_{:=\gamma_\lambda}$$
Furthermore, we know $n$ (total sample size)so:
$$E[X_j\mathbb{I}_{x^T\Sigma^{-1}x>\lambda}(X_j)]\approx \frac{1}{n}\sum_{j=1}^nX_j\mathbb{I}_{x^T\Sigma^{-1}x>\lambda}(X_j)=\frac{N_Y}{n}\frac{1}{N_Y}\sum_{j=1}^nX_j\mathbb{I}_{x^T\Sigma^{-1}x>\lambda}(X_j)=\frac{N_Y}{n}\overline{Y}$$
where $N_Y:=\#j:X_j^T\Sigma^{-1}X_j>\lambda$. So we have that
$$E\bigg[\frac{N_Y}{n}\overline{Y}+X_j\mathbb{I}_{x^T\Sigma^{-1}x\leq\lambda}(X_j)
\bigg]=E[X_j\mathbb{I}_{x^T\Sigma^{-1}x>\lambda}(X_j)]+\gamma_\lambda=\mu$$
which is our estimator. It remains to compute $\gamma_\lambda$ in some way. We have
$$E[X_j\mathbb{I}_{x^T\Sigma^{-1}x\leq\lambda}(X_j)]=E[X_j|X_j^T\Sigma^{-1}X_j\leq\lambda]P(X_j^T\Sigma^{-1}X_j\leq\lambda)$$
We can do
$$P(X_j^T\Sigma^{-1}X_j\leq\lambda)\approx \frac{n-N_Y}{n}$$
So consider
$$\frac{N_Y}{n}\overline{Y}+\frac{n-N_Y}{n}\beta$$

Here we may want to solve the penalized problem
$$\min_{\beta}\bigg\|\overline{Y}-\frac{N_Y}{n}\overline{Y}-\frac{n-N_Y}{n}\beta\bigg\|^2_2+\nu\bigg(\frac{n-N_Y}{n}\bigg)^2\|\beta\|^2=\min_{\beta}\bigg\|\frac{n-N_Y}{n}(\overline{Y}-\beta)\bigg\|^2_2+\nu\bigg(\frac{n-N_Y}{n}\bigg)^2\|\beta\|^2$$
We have
$$\bigg(\frac{n-N_Y}{n}\bigg)^2(\beta^T\beta-2\beta^T\overline{Y}+\overline{Y}^T\overline{Y})+\nu\bigg(\frac{n-N_Y}{n}\bigg)^2 \beta^T \beta$$
we solve by taking the derivative and setting it to zero, obtaining
$$\beta=\frac{\overline{Y}}{1+\nu}$$
Example:

